text:
Would it take 7 to 10 times more energy to lift a turbofan-helicopter?
I want to figure out how much energy it would take to lift a turbofan-helicopter.
For reference I started with a rotor helicopter and assumed $T=100\ \mathrm{kN}$ gross weight and $d = 16\ \mathrm m$ rotor diameter (compare S92 or AH64).
Using the propeller-formula
$$
T = ( 0.5*P^2*\eta_\text{tot}^2*\pi*d^2*\rho_\text{air})^{1/3}.
$$
with $\eta_\text{tot}^2=0.58$ (source) I get an engine-output of $P=1\,870\ \mathrm{kW}$
If I now take a typical turbofan and a take-off $tsfc$ of $10\ \mathrm{g/(kNs)}$ I get a fuel consumption of $1\ \mathrm{kg/s}$ for my $100\ \mathrm{kN}$ helicopter. If I furthermore assume 30% overall efficiency and $E_\text{Jetfuel}=43.5\ \mathrm{MJ/kg}$ I get an energy demand of $13\,050\ \mathrm{kW}$ to run the fan (neglected all the other stuff taking energy from the shaft).
Does this mean it would take 7 times the amount of energy to lift a turbofan-helicopter (with engines running at full throttle)? Or a magnitude if I assume cruise tsfc...
Does this suit with your experience? Or is there a big error/misconception in my short calculation?

Yes, that's precisely the reason why vertical take-off in jets is limited to fighters, which have insane engine power for their weight, and very rare even there.

Helicopter rotors are very large precisely for that reason. Lift from a rotor (or thrust from a propeller or fan) is obtained by accelerating a given mass of air m to a velocity v. The thrust obtained is proportional to the increase of momentum ∆mv. Obviously, you can get the same ∆mv by accelerating a given mass of gas to a high speed (propeller/fan) or by accelerating a larger mass to a lower speed (rotor). But the energy needed is proportional to v squared... Hence, and for the same thrust/lift, it's much more expensive, in energy or power terms, to use a propeller/fan than a big rotor ...

Using this logic, would it make sense to build planes with massive rotors (maybe retract/move them when near the ground)?

The Osprey V-22, when in airplane mode, has very big propellers, and flies quite well...

The complication is that an airplane's propulsion system faces an already moving (relative to the airplane) air stream. Big props work well at low speed but not so well as the aircraft gets faster. AIUI the ospreys proprotors are a compromise, they are larger than ideal for airplane mode and smaller than ideal for helicopter operation.

@peter green The size is also infuenced by the US Navy's storage requirements.

@xxavier your answer explains exactly why the AeroVelo Atlas human powered helicopter uses four huge rotors (the aircraft is larger than a 737) with low speed.

A turbofan helicopter must also take off by producing at least the thrust equal to the TO weight. With turbofans, the static thrust at zero speed is very much determined by the bypass ratio. You're mentioning a typical turbofan, so let's take a CFM56-3 with a bypass ratio of 6, this is a 100 kN engine.

The UH-60 has a max. TO weight of 22,000 lb and a rotor diameter of 16.4 m, so this would be your target helicopter. Its installed max. TO power is 3,086 SHP = about 2,300 kW. It needs excess power to climb, so I would say your required engine power is in the ballpark.
The CFM56 has a mass flow rate through the gas generator 53.5 kg/s (source an old textbook), so total mass flow = (1+6)*53.5 = 374.5 kg/s. Average exhaust velocity $V_e$= T / $\dot{m}$ = about 265 m/s. Power = $\frac {1}{2}\cdot \dot{m} \cdot {V_e}^2 $ = 13.2 kW

So I get roundabout the same numbers that you get: your statement is valid, a helicopter powered by a CFM56 engine takes about 7 times the installed power to lift off.

From dimensional considerations, the power $P = k\cdot T^{3/2}\cdot D^{-1}\cdot \rho^{-1/2}$. Hence, for the same thrust and density, $P = k\cdot D^{-1}}$ where k is some constant...

Try a Rolls Royce Pegasus rather than a CFM56!

@BrianDrummond Royce Pegasus is of course a successful vertical lifter of airframes. It runs 81.1 kg/s through the gas generator and has a bypass ratio of 1:1.2. If we make the same calculation as above, we get a supersonic exhaust velocity, which can only be accommodated by a convergent-divergent exhaust pipe. This the Harrier does not have, so the excess energy causes a pressure rise, which delivers thrust as well. But much more detail of the geometry of the Pegasus is required to properly make the computation. The CFM56 case allows this method because average exhaust speed is subsonic.

You face another problem. Helicopter rotors are also used to maintain control of the aircraft, by varying the pitch of the rotor blades on a portion of their arc.
With a ducted turbofan, you'd have to find another way to control it. Probably bleed air exhausted on the roll and pitch axes, which is how the Harrier and F35 do it when in hover.
So you would need to add to your power budget, enough reserve bleed air for control.


text:
How is the Max Zero Fuel Weight used in calculating weight and balance?
In some weight and balance apps there is an option to "Use Max Zero Fuel Weight".  Having never seen this in any of the aircraft I have experience flying, I am curious how this value is typically used in the calculation of an aircraft's weight and balance.
Can someone give me an example of how this value is used in a weight and balance calculation?

Maximum Zero Fuel Weight is an operating limitation intended to protect the wing spar structure by limiting the concentration of load, relative to all up weight, borne by the wings in the middle of the span.  In other words it limits spar bending loads at the wing roots imposed by the mass in the fuselage because any additional weight must be in the wing tanks where the additional load doesn't add to root bending.
Generically, Zero Fuel Weight can just mean all up weight minus usable fuel.  However, when an instruction says to use "Max Zero Fuel Weight" it's saying to apply a published MZFW value that is an operating limitation.  In your case you would enter the Max ZFW value from the airplane's Operating Limitations in the AFM into the app.

I think your answer would be stronger if you reworded to say that the MZFW is one of the structural design weights.

For pilot purposes, it's an operating limitation.

I would tend to agree with this answer.  The Learjet 35 has tip tanks.  It does not have a published max zero fuel weight but instead a max zero wing and tip fuel weight.  Learjet calculates the weight of the fuselage fuel as part of the latter just for the reason stated above.

"In other words it limits spar bending loads at the wing roots imposed by the mass in the fuselage..." In what situation this spar bending moment is more pronounced? In the air or when an aircraft is on the ground? Or perhaps both?

In the air, either from pulling G, or from a gust load. The more probable case is gust loads, say from flying too close to a thunderstorm cell. There is a "turbulent air penetration speed" and if you respect that and the loading limitations, you can't, in theory, overstress the airplane.

It is a limitation to calculate Restricted Take Off Weight.
RTOW is the lowest of the three :

OTOW(Operational Take Off Weight) - which is essentially the lower of your i)Max Certified Take Off Weight and the ii)Performance Limited Weight for take-off.
MLW(Max Landing Weight) + Trip Fuel
MZFW(Max Zero Fuel Weight) + Total Fuel onboard at Take-Off.

There could be two interpretations of this - Do we want to know (i) the Maximum ZFW for the particular flight, or do we want to know (ii) the calculations based on the MZFW certification, let's call it 'DESIGN' Limitation?
Here are figures for a large existing airplane:
MAX DESIGN WEIGHTS based (TONS)
DESIGN TAKEOFF - 351.5
DESIGN LANDING  - 251.2
DESIGN ZFW  - 237.7
The above weights must not be exceeded.
Typical DOW^^ - 175.0
^^ DOW - The Operator publishes the Dry Operating Weight. This is the basic/empty weight of the airplane alongwith standard furnishings, crew complement with their luggage, pantry load, etc. - generally any regular load that is not fuel and not payload. DOW is arrived at by calculating the sum of all components of the airplane, as well as actually weighing the airplane.
Example for interpretation (ii) above:
Using MAX DESIGN ZFW
MAX Payload (DESIGN) = 237.7 - 175.0 = 62.7 TONS
MAX DESIGN ZFW, limits Fuel to to 351.5 - 237.7 = 113.8 TONS
The Fuel tank capacity is about 145.0 TONS, so we lose a possible 31.2 TONS of Fuel/about 4hours of flight if we carry MAX DESIGN Payload.
An example of interpretation (i):
Based on Route, forecast winds, weights (!) etc. and available RWY at a particular point in time at a particular airfield, the dispatcher/Pilot calculations yield the following numbers:
RTOW = 342.0 TONS, REGULATED TOW, calculated for actual conditions
MINIMUM FUEL = 119.4 TONS, by calculation under actual conditions
RTOW - MINIMUM FUEL = MAXZFW = 222.6 TONS (= a loss of 15.1TONS)
i.e. only 47.6 TONS OF Payload compared to the MAX 'DESIGN' value of 62.7 TONS.
SOME ARITHMETIC
The Take Off Weight is the sum of the following three weights:
DOW (as described earlier at ^^)
PAYLOAD - The Commercial Dept. are the ones who know how much Payload there is for the flight. We assume, in general, that they always want to load the flight to the fullest.
FUEL - The Dispatcher and Pilot are the ones who calculate how much Fuel is required to despatch the flight.They also calculate the performance limited MaxTOW for the flight (aka Regulated TOW, RTOW).

Take Off Weight, TOW = DOW + Payload + Fuel

When we add Payload to the DOW we get the Zero Fuel Weight:

DOW + Payload = ZFW, this should not exceed the Max DESIGN ZFW Limitation in any eventuality.

Substituting  ZFW for DOW + Payload:

TOW = ZFW + FUEL.

As mentioned earlier, the Dispatch and Pilot calculate the Minimum Fuel and the Maximum TOW (RTOW)for the flight. The Pilot may choose to carry more than the Min fuel if the preflight briefing indicates a need for more fuel (marginal weather, last minute technical issues, etc.), last minute issues can also reduce the RTOW (RWY change eg). So  the allowable load for the day is:

RTOW - Fuel = Max ZFW for the flight.

This Max ZFW tells the Commercial Load Control office how much Payload can be loaded. This is the answer to interpretation (i).
For interpretation (ii) A "Use Max ZFW" option on the Performance App could be an option for relatively short legs where the fuel requirement is known to be well below any limits and there is an unsatisfied demand for Payload. Such a figure, is, in any case, a fixed figure being the DESIGN MZFW less the DOW:

DESIGN MZFW - DOW = MAX DESIGN PAYLOAD which should never be exceeded, however favorable the Fuel and TOW maybe.

Note that this is a fixed figure save for minor variations of DOW due things like variations in standard crew numbers, fewer galley carts, etc.

Zero Fuel Weight is the weight of the aircraft with everything loaded for that flight, minus useable fuel (Takeoff Weight minus weight of useable fuel).
The Zero Fuel Weight in your weight and balance calculations are just a reference mark. It is not used for much if anything in most flight applications since you should never get to within 30 minutes of your empty status. This reference is useful in calculating other reference points. One of which is your maneuvering speed (Va). If you know your Va at your takeoff weighting your Va at your Zero, it gives you an envelope to remain within when it is necessary to slow down to Va. Slowing down to your Landing Weight Va speed during turbulence is your best option.
Your Center of Gravity calculation is just as important. Keeping Takeoff, Landing, and Zero Fuel CoGs in the envelope ensures best performance for things such as spin recovery and stability during all phases of flight.

The question isn't asking about ZFW, but about Max ZFW, or MZFW.

Also, ZFW is used plenty for everything than shows fuel remaining in lbs or kg. Add current (or projected landing) fuel to your ZFW to get current GW or projected landing weight. About anything with an FMC has that functionality.

source: Stack Exchange

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text:
What currency requirements exist specifically for Commercial and ATP pilots that a Private pilot wouldn't know about?
Besides a different medical certificate, what other currency requirements do commercial and ATP pilots have to contend with that a private/instrument pilot would not?

§61.58   Pilot-in-command proficiency check: Operation of an aircraft that requires more than one pilot flight crewmember or is turbojet-powered.   https://www.ecfr.gov/cgi-bin/retrieveECFR?gp=&SID=65261edeb865e8e4e6001e6495f63529&r=PART&n=14y2.0.1.1.2#se14.2.61_158

Which country or regulations are you asking about?

Why would private pilots not know about such requirements (if they're interested, of course)?  It's not like they're kept secret until you're inducted into the society :-)

@Pondlife United States primarily ... but I'd be curious if there is anything outside as well.

Autoland currency
If equipped with autoland, the flight crew will have some requirement to use the system, or else autoland operation in real CAT III weather, where autoland is required, will be prohibited. For many types, this also applies to the aircraft itself.
Currency is satisfied by conducting a successful autoland in any weather, so in periods of good conditions, autolands may be practiced during normal CAT I or CAT II approaches. Crews can also do this in the simulator. All autoland use is reported to the company so they can track currency.
It varies by airline policy and aircraft type, but typical requirements may be one every 30 days, one every 90 days, or 2 or 3 landings between 6 month sim checks.
Some airlines will put a movable placard on the flight deck as a reminder to crews what the status of the aircraft is, e.g. you will see "CAT II" or "NO AUTOLAND" when the system is faulty or not current.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
What is the difference between a Mechanic and a Repairman?
In reviewing some of the data published by the FAA, I noticed that they make a distinction between a mechanic certificate and a repairman certificate.  What is the difference?

The basic difference is that a mechanic can perform maintenance on any aircraft; a repairman can perform maintenance only on aircraft for their employer, or that they own.
The full details are in 14 CFR 65 Subpart D (mechanics) and Subpart E (repairmen). Here are the basic privileges for each one (emphasis mine):

A certificated mechanic may perform or supervise the maintenance,
  preventive maintenance or alteration of an aircraft or appliance, or a
  part thereof, for which he is rated (but excluding major repairs to,
  and major alterations of, propellers, and any repair to, or alteration
  of, instruments), and may perform additional duties in accordance with
  §§65.85, 65.87, and 65.95.

And:

A certificated repairman may perform or supervise the maintenance,
  preventive maintenance, or alteration of aircraft or aircraft
  components appropriate to the job for which the repairman was employed
  and certificated, but only in connection with duties for the
  certificate holder by whom the repairman was employed and recommended.

"Repairman" also applies to experimental and light sport aircraft owners. If you build your own aircraft you have repairman privileges to maintain it, but you can't use those privileges to work on other aircraft, per 65.104(b) (emphasis mine):

The holder of a repairman certificate (experimental aircraft builder)
  may perform condition inspections on the aircraft constructed by the
  holder in accordance with the operating limitations of that aircraft.

This page from the FAA discusses both
https://www.faa.gov/mechanics/become/basic/

I'm a US Citizen. What requirements must I meet to get a mechanic's certificate?
You must be at least 18 years old;
able to read, write, speak, and understand English.
You must get 18 months of practical experience with either power plants or airframes, or 30 months of practical experience working on both at the same time.
As an alternative to this experience requirement, you can graduate from an FAA-Approved Aviation Maintenance Technician School.
  You must pass three types of tests;
a written examination
an oral test
a practical test
How do I get a repairman's certificate?
To get a repairman's certificate, you must be recommended by a repair station, commercial operator, or air carrier. You must
be at least 18 years old;
be able to read, write, speak, and understand English
be qualified to perform maintenance on aircraft or components
be employed or a specific job requiring special qualifications by an FAA-certified Repair Station, commercial operator, or air carrier.
be recommended for the repairman certificate by your employer
have either 18 months practical experience in the specific job or complete a formal training course acceptable to FAA.

So they look similar, with one having testing to go with it.

Also with an A&P license, once you have it, you do not have to get retested nor does it expire. The only way to loose the license is if you personally give up it. However it can be revoked.
A repairman looses their certificate once they leave the repair station.


text:
What might the business model be for a hypersonic jet?
According to this video, NASA and Boeing are doing research into a new generation of hypersonic passenger planes that can travel at Mach 5. The designs look really cool and I would love to see them in the air someday.
What I am curious about is the business model for such a plane. Are they planning on targeting high-end business travelers alone, or could the ability to make multiple flights in a very short time span bring the cost down enough for the rest of us to be able to afford a seat? Or is this another bid for the high-end market alone?
As a side note, whatever the market is for superfast travel, Elon Musk also seems to think that that there is a business opportunity there.

Learning from history, this might just be another subsidy for military research using civilian money. If there is no business case for supersonic travel, how can there be one for hypersonic travel?

I've built a semi-empirical model to evaluate a similar question for Mach 1.1-1.7 range, so plugged in M5 for a joke. The "super best case" scenario suggests $2,000+/pax just in fuel for JFK-HTR. The problems are thrust at takeoff, which with scramjets' low T:W makes for heavy engines, and long climb burning most of the fuel for the flight.

Well it’s cool but it’s nothing new.  Hypersonic transports and runway to orbit vehicles have been proposed since the late 1970s but never materialized into flyable hardware for a number of reasons, mainly engineers have not been able to solve the technical challenges associated with structures, propulsion, mission profiles, heat transfer and protection, operating costs for skittish airline operators, etc.
The new Boeing hypersonic transport may be the quantum leap forward for air and space travel but it could easily join the heap of fantastic ‘paper airplanes’ that never flew or possibly flew as clandestine prototypes such as:

Rockwell National Aerospace Plane - “Orient Express” from Reagan’s
1986 State of the Union Address.
Lockheed SR-91 “Aurora” reconnaissance aircraft.
High Speed Civil Transport concept from the
1990s
Lockheed Martin X-33 / VentureStar SSTO concept
German Sänger concept aircraft
Lockheed SR-72 Spy plane of the 2010s

As far as the passenger world is concerned, the big market are the transpacific routes, where the high-speed would make a major difference in travel times for both the public as well as business travel. If you could design a hypersonic aircraft that has a 6000 - 7000 mile range with operating costs in the neighborhood of 30 to 40% higher than existing subsonic airframes, you would have a game changer. But that’s always been the challenge.
Who knows, maybe the technology now exists to attempt a successful hypersonic manned aircraft.  I for one would welcome it as I find current aircraft are becoming insufferably boring.  OEMs churn out one bland subsonic twin jet after another, just wringing a few more knots out of a few less gallons of gas with new avionics etc.  I was supposed to be living on one of those cool Gerry O’Neill space colonies at this point!

The problem with high speed air travel is that it doesn't do anything about all the time you're not travelling at high speed: the hours spent going through check-in & security, boarding, taxiing, possibly going through customs & immigration, waiting for your luggage...

For short commuter routes yes. However for an LAX to Sydney route, flying at Mach 6-8 and making the trip in an hour and a half versus 17 hours in a conventional sub sonic aircraft that does take a whack out of the travel time.

But how much demand is there for those really long routes?  Even with say New York to London, ground time can easily be more than flying at cruise speed time.

BIG demand.  Do you want to sit in an airline seat for 8-20 hours or 1-3 hours?  You take your pick.  People will pay good money (but not excessive) for the latter.


text:
During an instrument approach, what are the obstruction clearance given?
I was wondering wether each segments of the approach has different minimum obstruction clearance? (E.g. initial, intermediate and final) From both terrain and man made obstacles.
If possible i would like it to be from ICAO, hence PANS-OPS. But any TERPS answers i would happily accept.

Each segment of the approach has different MOC (Minimum Obstacle Clearance). Below I have quoted the MOC from ICAO DOC 8168 Volume II for each of the different segments of the approach.
Initial segment (Part I - Section 4, Chapter 3.3.4):

The obstacle clearance in the initial approach primary area shall be a minimum of 300 m (984 ft). In the secondary area,
  300 m (984 ft) of obstacle clearance shall be provided at the inner edge, reducing linearly to zero at the outer edge.

Intermediate segment (Part I - Section 4, Chapter 4.3.2.1):

A minimum of 150 m (492 ft) of obstacle clearance shall be provided in the primary area of the
  intermediate approach segment. In the secondary area, 150 m (492 ft) of obstacle clearance shall be provided at the
  inner edge, reducing to zero at the outer edge.

Final segment, for non-precision straight-in approaches without vertical guidance (Part I - Section 4, Chapter 5.4.3.1):

The OCA/H for a straight-in, non-precision approach where the angle between the track and the extended runway
  centre line does not exceed 5 degrees shall provide the following minimum obstacle clearance (MOC) over the
  obstacles in the final approach area:
  a) 75 m (246 ft) with FAF; and
  b) 90 m (295 ft) without FAF.

Note that precision approaches and approaches with vertical guidance (ILS, LPV, LNAV/VNAV etc.) do not have a fixed MOC, and the obstacle clearance is calculated using more advanced methods.
Initial missed approach segment (Part I, Section 4, Chapter 6.2.1.3):

Obstacle clearance in the initial phase. In the initial missed approach area, the minimum obstacle
  clearance shall be the same as for the last part of the final approach area except where the extension of the intermediate
  missed approach surface backwards towards the missed approach point requires less clearance. (See Figures I-4-6-4
  and I-4-6-5.)

Intermediate missed approach segment (Part I, Section 4, Chapter 6.2.2.3.1):

In the intermediate missed approach phase, the minimum obstacle clearance shall be 30 m (98 ft) in the
  primary area, and in the secondary area the minimum obstacle clearance shall be 30 m (98 ft) at the inner edge,
  reducing linearly to zero at the outer edge.

Final missed approach segment:

In the final missed approach phase of a straight missed approach the minimum obstacle clearance shall
  be 50 m (164 ft) (Cat H, 40 m (132 ft)) in the primary area, reducing linearly to zero at the outer edge of the secondary
  area.

Figure I-4-6-4 and I-4-6-5:

The FAA lays this out in Order 8260.3c the three main relevant sections for requirements, level flight and climbing/decent segments are:

2-1-2. Required Obstacle Clearance (ROC). This order specifies the minimum measure of obstacle clearance considered by the FAA to supply
  a satisfactory level of vertical protection. The validity of the
  protection is dependent, in part, on assumed aircraft performance. In
  the case of TERPS, it is assumed that aircraft will perform within
  certification requirements.

...

2-1-3. Level OCS. The level OCS concept is applicable to “level
  flight” segments. These segments are level flight operations intended
  for en route, STAR, feeder, initial, intermediate, and NPA final
  approach segments. A single ROC value is applied over the length of
  the segment. These values were determined through testing and
  observation of aircraft and pilot performance in various flight
  conditions. Typical ROC values are: 1000 feet (2000 over designated
  mountainous terrain) for en route, STAR, and feeder segments, 1000
  feet for initial segments, 500 feet for intermediate segments, and
  250-500 feet for final segments.

...

2-1-4. Sloping OCS. The method of applying ROC, in segments dedicated
  to descending on a glidepath or climbing in a departure or missed
  approach segment, requires a different obstacle clearance concept than
  the level OCS because the ROC value must vary throughout the segment.
  The value of ROC near the runway is relatively small and increases
  throughout the segment.

The document is lengthy and lays out various formulas and charts, to keep form copying the whole thing here please reference the document directly for anything more specific.


text:
What does the letter "D" next to the VOR frequency and name mean?

In most charts there is the letter D next to the VOR freq. What does this mean? I assume the H and L in brackets in other charts mean high and low VORs (correct me if im wrong). But what does the D itself stand for?

Since you're specifically referencing a Jeppesen chart, we'll go right to the source.
Jeppesen's "Introduction to Jeppesen Navigation Charts" has a glossary and explanation of all terms and symbols.
The answer to your question reads as follows:

1 — VORDME. Low and High/Low charts include a Compass Rose with VHF
  Navaids. Shadow box indicates navaid is airway component, with
  frequency, identifier, Morse code and INS coordinates. Small "D"
  indicates DME/TACAN. Class indicated by: (T) Terminal, (L) Low, (H)
  High.

So, the D indicates that the VOR has a co-located DME radio.
(and, no need for correction to your assumption: the H and L in brackets do, indeed, refer to a High or Low class of VFR)


text:
Y-20 over Frankfurt
Saturday was a super clear day south of Frankfurt, Germany, near the airport (maybe 10 km away) and I saw a giant 4 engine plane flying over head - lit up by the sunset. It caught my eye because of the very broad wing, then with binoculars I could see it had a large red-ish oval strucure underneath it, a T-tail, and strut-like protusions on the wing bottoms. Googling silouettes afterwards I came to the conclusion it had to be a Chinese Y-20 but....
There are only like 20 of them operated by the PLA, and as there are no airshows going as far as I can tell, I am completely surprised. So the question is, why would it have been there? Or maybe I mistook it for something else?
I wasn't quick thinking enough to take a photo, though it would have been small and barely visible, it would have been something.

There are several transport aircraft with similar layout including C-5 Galaxy, C-17 Globemaster and Il-76. The last is notably operated by some civil operators as outsize cargo hauler.

You sure it wasn't an Emirates A380? Giant plane, red belly, and much more likely to be found at Frankfurt...

Ramstein Air Base, one of the largest US bases in Europe, is not too far away from Frankfurt. Probably it was a US military transporter (C-5, C-17).

Can you narrow down the time and location? That might allow finding the aircraft over on https://www.flightradar24.com/ (past 7 days are available free, longer history for subscribers only).

Sunset was approximately at 18:47 local time. Around that time an Emirates A380 passed over the airport enroute from London to Dubai, course west-northwest to east-southeast. But there are other candidates. Please add details such as time and location of spotting, observed course etc. to your question.

@MikeWise I'm also 10 km south of FRA (Gräfenhausen) and A380's fly over several times a day, typically (due to my location) on a south-eastward departure route. Are you a regular skywatcher in this area - would you know for sure that it wasn't an A380? Naturally I'm curious now too. :) I hope you solve it!

From below, the Y-20 and C-17 are near identical.

Y-20 from https://nationalinterest.org/blog/reboot/chinas-y-20-real-china-military-plane-you-need-fear-185509
C-17 from https://www.aerotime.aero/28603-800-people-board-c-17-flying-out-of-overrun-kabul-airport
Ramstein AB (USAFE) being a short hop from Frankfurt, that would be the logical conclusion.

Yeah, somehow my Google-fu failed me

@MikeWise probably because you included the word "red" it got resolved into "red China"

Also thinking the reddish-ness may have come from the sunset lighting, though it seemed to be painted on. What is that structure called? Seems like a lot of transports have it.

What 'structure'?

It's a possibility the aircraft was a C-17, but we don't know. That should be a comment until you have more information, like a reference form a ADS-B tracking site.

Good answer. You could probably improve it by selecting a C-17 image with a red stripe instead of a green one on the tail, by mentioning he other operators, or by pointing out that there's a lot of room for squadron colors. (examples: https://live.staticflickr.com/5161/5300236538_a412fec774_n.jpg , https://c1.staticflickr.com/5/4262/35467031100_1cffa36bcf_b.jpg )

@MikeWise landing gear fairings? (I assume you are asking about the lumps the wheels are poking out of on the Y-20)


text:
Which AFDX connections are used in Airbus A380?
AFDX is defined by ARINC 664 Part 7 and used in Airbus A380, I would like to know in which connections on the aircraft is used and for which purposes? Does anyone knows the answer or good documentation from which to start a research about?

Can you clarify what you mean? Are you asking what type of devices are connected together on the aircraft's network via A664?

@scotty3785 I really mean any kind of connection, the point is I'm interested in study its usage on Airbus A380 and I will need to understand where is used. An example of answer I could expect it would be "It's used in the IFE infrastructure to connect LRU of the seats togheter or with the IFE Rack Server". I hope this explained a little bit my question.

I haven't seen any reference to the IFE system. Common sense would dictate that they systems would be partitioned from each other but I don't know for sure.

There are some resources.
This source shows how AFDX is used as the backbone of the IMA (Integrated Modular Avionics) and lists subsystems such as the air data inertial reference units (ADIRU), Engine Electronic Controllers (EEC) and Electrical Load Management (ELM) Systems as 'members' of the network.
This archived article talks about the development of AFDX to solve real time performance issues with the flyby wire controls of the A380.
Other articles of note would be this. The other articles are behind a paywall but for completeness they are here and here

Please send me also the link of the articles behind pay wall, in case of need is better to have them and I will decide if they worth the value. Thanks.
Of course I mean the names of the articles in case you cannot send to me the effective links.

@Bluephoenix I've added the links that I was able to find.

@Bluephoenix You're welcome.

I finally add time to go through all the documents, one of them in the wayback machine is actually empty and the sources are unreachable, the other one are really useful and well made though they talk about the IMA and AFDX but not really mentioning where actually is used on an Aircraft.
Have I to suppose that in case AFDX is used together with IMA Architecture it connect everything and no other ARINC protocols are used in the Aircraft Network?

For example there is something in the Passenger Domain that is connected through the AFDX ( for example: SDU, FDB or IFE rack/terminal/server).

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Pilot seat - primary or secondary structure?
For design of pilot seat, would the seat be considered as primary or secondary structure?

I wouldn't consider it structure as most are designed to be movable on some kind of rail system.

How does that disqualify it from being a structure? It still needs to sustain loads, not mentioning that you would normally not want your seat to be moving during operation of the aircraft.

I can see that.  I would suggest Primary then, as the Pilot seat at least is usually in  a fixed location prior to takeoff. (I never move mine, anyway).

In what kind of context are primary and secondary structure defined? Please add a reference, as the definition of primary and secondary structure may greatly vary depending on the context.

Seats are not structure. They are equipment.
I believe you mixing system/sub-system definitions with design requirements. Primary and secondary structures are structural systems within the aircraft that have a primary or secondary purpose of transferring a load. Everything else on the aircraft is classified according to its primary function (air conditioning, flight controls, etc.)
Having said that, there are structural requirements that exist for all the components of these functional systems. That doesn't make them structure.
To illustrate the point, we can refer to ATA spec 2200 which provides a common structure for aircraft documentation. The documents use use a Chapter-Section-Subject numbering system that correlates to a System-Subsystem-Subject.
At the highest level, Chapters 20 through 50 cover Aircraft Systems, 50 - 59 cover structure, 60 - 67 cover Propeller/Rotor, and 71 - 84 cover Power Plant. Looking at the Chapter definitions we find (emphasis mine):

Chapter 25,   EQUIPMENT/FURNISHINGS
Those removable items of equipment and furnishings contained in the
  flight and passenger compartments. Includes emergency, galley and
  lavatory equipment. Does not include structures or equipment assigned
  specifically to other chapters.

And under that is:

Section -10,  Flight Compartment
The compartment above the floor and between the forward passenger
  partition and the forward pressure dome. Includes items such as flight
  crew seats, tables, pilot check lists and food containers, wardrobes,
  curtains, manuals, electronic equipment rack, spare bulbs, fuses, etc.
  Does not include cargo compartments.

Further along, we find:

Chapter 51,   STANDARD PRACTICES AND STRUCTURES - GENERAL
Standard Practices, General Procedures and typical repairs applicable
  to more than one chapter and are not specifically covered under
  [Chapter (System) 52] thru [Chapter (System) 57] Sub-Sys/Sect
  breakdown.
Section -10,  General
Airplane major structural breakdown diagram. Primary and secondary
  structure diagram. Principal area and dimensional data. Restricted
  area diagram. Zoning diagram. Access door and panel identification.
  Glossary.

The structural area of interest here is:

Chapter 53  FUSELAGE
Structural units and associated components and members which make up
  the compartments for equipment, passengers, crew, cargo, plus the
  structure of the envelope and gondola of airships. Includes skins,
  belt frames, stringers, floor beams, floor, pressure dome, scuppers,
  tail cone, fuselage-to-wing-and empennage fillets, attach/attached
  fittings, load curtains, cables, ballonets, etc.

You'll note that this specifies only physical structure, not systems or equipment installed in the fuselage.
To address the structural design requirements, we can refer to 14 CFR Part 25, Airworthiness Standards: Transport Category Airplanes.
Subpart C with the title Structure is broken down into the loads the aircraft must be designed to withstand (flight loads, gust loads, ground loads, etc.)
Subpart D covers Design and Construction which provides specifics for individual design elements of the aircraft. Therein we find:

§25.785   Seats, berths, safety belts, and harnesses.

Within this section is the necessary design requirements for seats, etc. which reference the applicable load specifications of Subpart C. A detailed reading of this section will reveal that the design loads for the seats are driven by the Emergency Landing Loads defined in Subpart C.

That falls under the subject of Flight Deck and Furnishings. The Airline Transport Association (ATA) classifies that equipment under Chapter 25 Equipment and furnishings.  OEMs as well separate that equipment from structural airframe components as well.  Primary and secondary structures refers to structural components in an airframe.
For example, a fwd wing spar or a fwd pressure bulkhead are primary structures.
A Douglas clip might be an example of a secondary structure, etc.

I disagree, it is not subject to flight deck (especially since pilot seats are in all aircraft that do not have flight decks), and it is not furnishing. The only way you can get out of the question through semantics, is by saying it is a safety equipment. But that would not help either.

If you're going to disagree with everyone why ask the question? You haven't even given a context. Anyway, it seems pretty obvious that there's nothing intrinsic in your average pilots seat that relates to the flyability of the aircraft so I don't see how it could be considered structural. A gauge wouldn't be a Primary or Secondary structure either

"if I am going to disagree"? You are actually just avoiding the question.
Right, so an example:
An aerobatic aircraft goes into a manoeuvre, the back support bends, moving the pilot away from controls. Not being able to unbuckle, the aircraft crashes, as a result of structural failure.
That would be my argument for it being primary structure. However, if i have to provide arguments myself, why the hell would I be here?

Uh, no.  That’s how Boeing and most OEMs classify that equipment.

@LuTze Just because something is critical doesn't mean it's structural. The plane will crash if the Pilot has a heart attack, too - but it doesn't mean they've become structural!


text:
Why bother tracking TSN after an overhaul?
I've noticed that mechanics often like to keep track of "time-since-new" (TSN) on components that have already been overhauled.  What would TSN matter when the component is tracked by TSO for maintenance purposes?

Lets you track how many times it's been overhauled, for one thing.

@ZeissIkon Do you happen to know of any component life limits that put an upper limit on the number of times it can be overhauled?

@CEOofAeyrium some compressor blades have an absolute life limit due to fatigue cracks that can form.

No, if the part is not subject to a life limit of its root component.
In the airline business, there are actually two terms used in the industry - "overhaul" and "restoration" which can generally be defined as follows:
If a part doesn't have a structural life limit, then the part can be rebuilt to new specifications an indefinite number of times, and each time, it is equivalent to a new part and will generally be called "overhauled" or "zero timed".  Mechanics can track "time since new" but they don't really have to and if parts move around it probably won't be anyway.  Piston aircraft cylinders are like this.  If you buy a rebuilt cylinder, nobody will know how many hours are on it since it was first made.
However a lot of components, especially components that operate under stress on transport aircraft, can have structural life limits applied as a result of endurance tests (where, say, a series of housings under endurance test will start to develop cracks at X thousand cycles or hours, and a life limit will be applied at 1/3rd of the endurance result, which is the safety factor applied to structural cracks that develop in test).
When these parts get rebuilt, they aren't called "overhauled", they are called "restored".  A structural actuator may get a restoration interval applied where it has to be rebuilt (because parts wore out during endurance testing) but once the root component like the basic housing gets to the life limit, the entire part is scrap.  In these cases the rebuild is not an overhaul but a restoration.  The hours or cycles have to be tracked since new since once it hits the life limit it is scrap even if it was rebuilt (restored) the previous month (which an airline won't let happen).

Depending on the part we are talking about, namely system components or engine LRUs, it becomes really useful for failure statistics and reliability analysis to track TSN/TSR/TSO & CSN/CSR/CSO of such units, even though there may not be hard limits to their life.
Engine & systems OEM maintenance programs usually contain recommendations on "soft times" and scheduled repairs to mantain acceptable reliability, it is widespread standard practice to track these sort of parts (serialized) in a management software of some sort.


text:
Is there such a concept as 'Max Overhauls' for aircraft components?
I have never seen any overhaul-able item with a documented limit on the number of times it can be overhauled.  In all cases it was subject to analysis and predicated on it passing certain tests.  Has anyone else heard of such a limit?
If so, what examples of aircraft components with such limits and what kind of aircraft can it be found on?

Yes, but its a fairly new change from the FAA. The FAA recently regulated part Life Limits, a limit in place at certification time (or through a later AD etc.). The IATA also outlines guidance for life limited parts. This is a fairly new reg and many of the older aircraft out there may not be subject to it since they were certified prior to it. This AC also outlines some info relating to it.
However for any mechanical part there, in theory, is a limit to the amount of times it can be overhauled if that work requires reductive machining. Something like a crank case, cylinder head or piston sleeve can only be machined so many times before there is not enough material left for a proper overhaul.
The definitions of overhaul vs rebuild are important. A rebuild may reset a parts life, you can read more on that in this question.
Any life limited part should be listed in the documentation for the specific airframe/model in question. For example here is the list of the Robinson R-22 helicopter. More notably the Cirrus SR-22 has an airframe life limit of 12,000 hours as part of its certification.

The certification requirements of 14 CFR 23.573 require that the
  composite airframe structure, cabin, wing, empennage their carry thru,
  and attaching structure whose failure would be catastrophic must be
  designed to damage tolerant criteria. Damage tolerant certification
  for the selected airframe life of 12,000 flight hours has been
  established for all of the affected parts with no special structural
  limitations or inspections.

Hourly life limits on components is fairly common ... what I haven't seen personally is the concept of a documented max number of overhauls allowed.  The example given where there is reductive machining makes sense, but that would be an example of an on-condition ... in my question I was wondering if there were any aircraft manuals that have a documented number of times a crank case can be overhauled (e.g. 5 times) or, as I would expect, is likely worded "can be overhauled as many times as the material of the crank case will allow".

@Aeyrium technically that could be stipulated (i dont know of any cases) but it does not make much sense from a mechanical/engineering standpoint. Wear and tear does not always occur at a linear rate and TBO times are really just suggestions so its most important to routinely watch components and base your overhaul decisions on condition.


text:
Under what circumstances can a flight attendant unbuckle and get up during landing?
Lately I've been on several flights where the flight attendants had to talk sternly to passengers that were getting up either just before landing or just after (which is quite obvious to everyone since the talking to is done over the PA system).
Now this made me wonder what FAs can do if the passenger just refuses to comply? Can FAs get up and forcefully seat passengers (would they have to talk to the pilot first to abort the landing)? If not, are there any circumstances where the FAs can/should/must unbuckle during a landing?

Are you asking about a specific country or set of regulations?

@Pondlife: no, I was just curious, so I'm happy with answers for any country really (the more the merrier )

An FA is a member of the crew and can get up or sit down as they wish during any phase of the flight. As to what they can do, they can have the pax arrested the moment the aircraft door opens. If that is not a good reason to obey instructions, I don't know what is! If the pax becomes a threat to the safety of the flight, yes, they can certainly get physical to put an end to the situation and secure the cabin.

Plenty of passengers have been arrested due to doing stupid things, so I'm assuming the case of a really stupid passenger who is hellbent on being stupid. Anyways, thanks for the answer!

The FA's are trained for that too. And remember, the likelihood these days is that there is at least one flight marshall on the aircraft, and sometimes more than one. Hellbent on stupid in that case will meet hellbent on "Meet my friends the thick nylon straps." The third point is that these days pax are no longer the sheep they used to be and will also get physically involved as well if necessary.


text:
How does the power transfer unit in the A320 work?
As far as i know the function of a PTU is to power one hydraulic system (either green or yellow) that is experiencing low pressure driven by pressure from the other system which still has the normal 3000 psi, where no fluid transfer happens. I have read (correct me if im wrong) that this happens with the help of hydraulic pumps, but how exactly does the "power" transfer happen?

It involves a lot of barking.

Yes it does @AEhere

The PTU consist of a hydraulic motor that drives a hydraulic pump.
The motor is driven by one hydraulic system, the shaft of the motor is connected to the pump and the pump powers the other hydraulic system.


text:
What is the width of a high altitude airway?
Victor airways (airways that are below 18,000 ft) has a width of 8 NM. How about those airways that operates higher than 18,000 ft, do they have specific width you must be in?
Or is there no width because you would be using GPS? (ones with an RNP of 2 NM)

At least in the US airspace definition, High Altitude Airways (Jet Routes) have no defined width regardless of the airway being navaid-based or RNAV/RNP
This is briefly stated in the FAA training manuals:
http://tfmlearning.faa.gov/Publications/atpubs/AIR/air2004.html

There are several Navigation Performance Minima defined by ICAO.
For example MNPS for the Atlantic Oceanic is 12,6NM later in 95% of time.
Since e.g. the Atlantic is a very bussy airspace, but without ground based Navigation Aids and communication to a controller is difficult, there is a incentive to improve the nav perf of aircraft.
There are several regions (Shannon, Gander, New York, Santa Maria, Reykjavik) of wich a at least one has already adopted a MNPS 10 instead of MNPS 12,6.
The goal is to eventually reduce it to MNPS 4, but that might take a while.
In addition to that there is also vertical seperation, which up to FL 285 is normally 1000ft (semicircular flight rule).
From FL 290 it is 2000ft, exept in RVSM airspace (reduced vertical seperation minima), where it is 1000ft up to FL 410.
Above that vertical seperation is always 2000ft.
So a lot of planes fit into a quite small volume. Planes which can not satisfy these criteria are not allowed to fly within them.
Since I'm in the process if learning this stuff, I might have missed something or got something wrong, but it might give you an idea.


text:
How does fuel flow and altitude affect the compression ratio of jets?
Does the compression ratio of jet engines change with altitude and/or with the rate of fuel flow?

Sorry about that ymb1. Thanks for the info

the compression ratio of a conventional gas turbine is a strong function of the rotating speed of the engine; note that at near-zero speed, the compression ratio of a gas turbine is essentially 1:1 and the engine will not run- even if some means is provided to alter the pitch of the compressor stage fan wheels.
density altitude also enters into the picture, inasmuch as beyond a certain altitude, there is not enough oxygen available to allow the engine to develop full rated power- and beyond that point the engine's power rating falls off with increasing altitude.

So basically the faster it spins the higher the compression ratio, and the slower it spins the lower the compression ratio. Yes?

yes; and the relationship is such that if you try to throttle the turbine below about 60% of max fuel flow, it almost stops running- which is why ground idle for a turbine is around 60% thrust.

@nielsnielsen, idle is usually indeed around 60%, but N2. Are you sure that means it is also 60% fuel flow and thrust? I really doubt that as while idle thrust is enough to get some planes moving, it will only get it moving at taxi speed. Also, while N2 is indeed always around 60%, the N1 is around 30% or even less for the newest high-bypass turbofans.

Related: https://aviation.stackexchange.com/questions/9526/what-is-the-relationship-between-thrust-and-engine-rpm-in-a-turbojet

@janhudek, my info is old, based on turbojets, and I cannot remember if the figures were for %speed or throttle setting or flow rate. recommend you post a better reply & I will delete mine. -NN

At high altitude the ambient air pressure is lower therefore the pressure ratio of the jet will rise. Since the compression occuring within the jet is adiabatic the compression ratio will also rise. Increasing the fuel flow rate will increase the working temperature of the jet and hence it's working pressure. Therefore again, the compression ratio of the jet rises.


text:
What is the definition of mountainous area?
MEA, MORA, grid MORA, and MOCA for a non-mountainous area gives 1,000 ft clearance from the highest obstacle, and in mountainous area gives 2,000 ft clearance from the highest obstacle.
But what is the exact definition of a mountainous area?
I have also heard that MOCAs have a different clearance calculation for the mountainous area depending on the terrain/obstacle elevation, (3000 ft - 5000 ft gets 1500 ft clearance) and (above 5000 ft gets 2000 ft clearance) can someone clarify this.

(I'm assuming you're asking about the USA.)
14 CFR 91.177(a)(2)(i) says (emphasis mine):

In the case of operations over an area designated as a mountainous
  area in part 95 of this chapter, an altitude of 2,000 feet above the
  highest obstacle within a horizontal distance of 4 nautical miles from
  the course to be flown; or [...]

If you then go to 14 CFR 95, you'll see that it defines mountainous areas by latitude and longitude. It's entirely useless for practical purposes, but at least it's precise :-)
As for the MOCA, I can't find anything to support the different definitions that you mentioned. 95.1(e) simply says:

(e) The MOCA assures obstruction clearance on an ATS route, ATS route
  segment, or other direct route, and adequate reception of VOR
  navigation signals within 22 nautical miles of a VOR station used to
  define the route.

The AIM and PC/G support that; there's no other definition that I could find.

Thanks alot for that :D

Designated mountainous areas include those areas having a terrain elevation differential exceeding 3,000 feet within 10 nautical miles within those one arc-second quadrangles overlying terrain or U.S. territorial waters. This new definition would also align with that used by ICAO.

Do you have any sources you can add to back up your answer?


text:
What are the duration of TAFs?
TAFs (terminal aerodrome forecast) have different durations. Some forecasts 30 hours into the future while others forecasts only 24 hours.
But what is the shortest and longest TAFs? and how often are TAFs updated?

First you will need to learn to decode the TAF's validity period.
TAF YMAY 022230Z 0300/0312 35010KT CAVOK
FM030800 31018KT 9999 SHRA BKN025 OVC100
INTER 0308/0312 31020G40KT 3000 +TSRA BKN010 SCT040CB
RMK FM030600 MOD TURB BLW 5000FT
T 23 24 28 33 Q 1012 1013 1014 1009
The validity of this TAF (in bold) is formatted as ddhh/ddhh.

Validity period of TAF is from 0000 to 1200, on the 3rd day of the
  month UTC

From here we can see that it's valid for 12 hours.
Another example:
TAF COR YMLT 212240Z 2200/2218 31015G28KT 6000 -RA BKN010 OVC100
TEMPO 2209/2218 2000 +TSRA BKN005 SCT040CB
RMK
T 25 21 18 15 Q 1014 1013 1013 1011

Validity period of TAF is from 0000 until 1800 on the 22nd of the month UTC

This one is valid for 18 hours.
You may check your country's AIP for how often they are updated.
For instance, in Australia, GEN 3.5 - 5 describes the routine TAF service based on airport size.

A International:  Issued 6 hourly, valid for 24 or 30 hours.
Commencement times 00, 06, 12 and 18 UTC.

B Large: Passengers > 150 000 or Movements > 75,000
Issued 6 hourly, valid for 12 or 18 hours. Commencement times 00, 06, 12 and 18 UTC.

C Medium:
Passengers > 50,000 or Movements > 10,000
Issued 6 hourly, typically valid for 12 hours.
Commencement times are 02, 08, 14 and/or 20 UTC, except in Western Australia where commencement times are 04, 10, 16 and/or 22 UTC.

D Small:
Aerodromes meeting passenger and movement thresholds, or other operational criteria.
Issued 6 or 12 hourly, valid for up to 12 hours.
Commencement times are typically 20 and/or 02 UTC, except in Western Australia where commencement times are typically 22 and/or 04 UTC.

Thanks for this additional info about Australia @shogunnyan. But i found what i was looking for in ICAO annex 3, part 6.2.6 nevertheless thank you for the answer. :D


text:
What is the visual manoeuvring area (circling approach) radius for PANS-Ops (ICAO)?
I'm very confused of the actual radius of the "protected area" in a circling approach for a PAN-OPS approach plate. The ICAO doc 8168 contains information for circling approach at 1 000 ft. But what about airports that are sea level? How do you calculate the radius?
P.s. Not TERPS :)


text:
Does the stator blade angle changes? Why?
I have read that when jets are on the ground their compression ratio changes. With this change the stator compressor's blade angle changes. Is this true?

The picture above shows the velocity triangles for one axial compressor stage. You can see that a big factor for the angle at which the airflow leaves the rotor, is strongly dependent on U the rotational speed of the rotor. So when the engine runs at different powersettings the air is gone come at the stators at different angels(since U is changing). Since the stators are like wings and have an optimal angle of attack, it makes sense to make them rotate so they can always keep this optimum angle, even if the oncoming flow changes its angle.
So it doesn't really have much to do with whether the plane is on the ground or not, the major factor for the stator adjustment is the changing rotational speed of the compressor rotors.

Where did you read this? Variable Inlet Guide vanes (VIGV's) have many different names across different engine manufacturers, but the principle is the same, a number of stator stages are operated mechanically to maintain compressor efficiency across the RPM range the engine operates dependant on altitude, demand etc, it doesn't strictly work on the ground.


text:
Vortex Lattice Method with eliptical panels
I implemented vortex lattice method (or deeper here) into my aircraft simulation program. The problem is that sometimes trailing vortex filaments from Horse Shoe Vortexes of main wing hits panels of rudder/elevator which leads to singularities (infinite coupling between the two panels). This is because rectangular panels generate sharp singular vortex filament at the end of wing.
(see https://scicomp.stackexchange.com/posts/30062/edit)
I'm thinking to use eliptical panels as a basis set for VLM to avoid those singularities - then I hope my simulator can work with smaller number of panels (I wan't to make it real-time).
This is normally ploted in texbook:

My questions:

I search for formula / derivation how the down-wash from eliptical wing looks like in the whole 3D space. Usually what is plotted in texbook (constant rectangular downwash) is true just immediately behind the wing.
Is it fine to express wing of complex shape as linear combination of eliptical wings and it's downwash as linear combination of downwashes from each of them? I mean if potential flow is linear problem, than I see no problem - ut maybe I forgot about some catch.

Interesting problem. I however don't really get the idea of elliptical panels. Your textbook figure approximates infinitesimally small panels, yielding an elliptical lift distribution. (There is probably a previous figure in which the discrete vortices / panels are still visible.)
Have you already taken a look at the (open source) codes of Drela's AVL and Tornado (by Tomas Melin)? Especially the latter takes a somewhat different approach in wake modelling, based on Eppler (1997), that might help/inspire you.

maybe "panels" is not propper word. I basically want to express down-wash as a linear combination of some nice smooth basis functions (rather than singular horse-shoe vortexes produced by rectangular panels). And one of the possibilities which comes to my mind is eliptical lift distribution, which produce constant rectangular down-wash functions.

The elliptical lift distribution is the result, but if you discretize that - you'll end up with some form of panels. Of course, the more you have, the more continuous the distribution's going to get, but if you want to use it for real-time applications, that's probably too computationally expensive. Again I point you to Tornado, which uses vortex slings instead of horseshoe vortices. These consist of 7 vortex filaments, and might provide some more flexibility. Also, I'm wondering if you can't prevent the singularity in numerically - e.g. adding some noise to prevent the vortices lining up.

I prevent the singularity now by using 1/(r^2+w^2) instead of 1/(r^2) in expression of vortex strenght. But then the vorticity is not stricly conserved, so I don't want to use too large w. Than the velocity field is not very smooth and I need rather fine panels to sample it properly.

ad The elliptical lift distribution is the result, but if you discretize that - you'll end up with some form of panels ... No, I don't want to discretize eliptical distribution using some other basis-functions (e.g. by panesl and correspedning horse shoe vortexes). I want single analytical basis-function describing analytically velocity field generated by the whole eliptical lift distribution. Than I want to express other lift-distrubutions by linear combination by such basis - with 1 expansion coef. assigned to each eliptical basis function.

thanks for "Thomas Melin" ... I downloaded tornado at some point by without background explanation I did not wan't to dig-in. But now I found this thesis with goodle scholar
https://s3.amazonaws.com/academia.edu.documents/30929514/thesis.pdf?AWSAccessKeyId=AKIAIWOWYYGZ2Y53UL3A&Expires=1535615224&Signature=JVHK4wuj1IlPfJt2OW3Q3u5W%2BEc%3D&response-content-disposition=inline%3B%20filename%3DA_Vortex_Lattice_MATLAB_Implementation_f.pdf

Typically in numerical simulations, instead of an ideal(inviscid) vortex, a viscous vortex with a finite core is used to avoid the singularity at the centre. In these viscous vortices, the swirl velocity inside the core is assumed to be linearly increasing from zero at the centre. While outside the core, the induced velocity remains the same as that of the ideal vortex.
There are multiple models that model these finite core vortices; some of which are:

Rankine
$$V_{\theta}=\begin{cases}\overline{r} & \text{if }\overline{r} \leq 1 \\ 1 / \overline{r} & \text{if }\overline{r}>1\,.\end{cases}$$
Lamb–Oseen
$$V_\theta=\frac{1}{\overline{r}}\left(1-\mathrm{e}^{-1.2526 \overline{r}^{2}}\right).$$
Bumham–Hallock
$$V_{\theta}=\frac{\overline{r}}{\overline{r}^{2}+1}\,.$$
Proctor

$$V_{\theta}=\begin{cases}
  \frac{1.4}{\overline{r}}\left(1-\mathrm{e}^{-\beta(1 / \overline{B})^{0.75}}\right) \times\left(1-\mathrm{e}^{-1.2526 \overline{r}^{2}}\right) & \text{if }\overline{r} \leq 1 \\
  \frac{1}{\overline{r}}\left(1-\mathrm{e}^{-\beta(\overline{r} / \overline{B})^{0.75}}\right) & \text{if }\overline{r}>1\,.
\end{cases}$$

Proctor–Winkelman

$$V_{\theta}=\frac{1}{\overline{r}}\left\{1-\exp \left[-\beta_{i}(\overline{r} / \overline{B})^{2} /\left\{1+\left[\left(\beta_{i / \beta_{o}}\right)\left(\overline{r} /_{\overline{B}}\right)^{1.2}\right]^{p}\right\}^{1 / p}\right]\right\}\,.$$

Vatistas
$$V_{\theta}=\frac{\overline{r}}{\left(\overline{r}^{2 n}+1\right)^{1 / n}}\,.$$

In these equations, $\overline{r}=r/r_\mathrm{core}$.


text:
Tail Wing Dented—Is it safe to fly?
I’m sitting in the lounge waiting to board this flight. Saw a dent in the attachment of tail wing to the body. Is this common?
Also I could be wrong and it is part of the design of the airplane.
Any guidance is appreciated!

Related: Why is the horizontal stabilizer root interface designed this way?

Thank you @fooot. Will read more into the design choice. Thanks!

Technically, this is called an adjustable stabilizer. https://en.wikipedia.org/wiki/Stabilator

Although the tail wing wasn't dented in this case, it would have been interesting if someone had answered the title question. IF it were dented, would it be safe to fly?

@Cloud no, we don't answer hypothetical questions.

@Cloud depends on how and where its dented

The aircraft in question is an Embraer ERJ 175 the "dent" allows the tail trim to operate properly you can see it more clearly in this picture.

(source)

So why in the OP was the trim seamingly set to hard 'up'? (Before takeoff)? I would inform the pilot.

@Cloud no, you would not inform them, because they have checklists to run (i.e., they know the setting) and they don't care about the setting on ground standing still at the gate.

@Cloud Unless I'm aware of 1) a checklist for this particular type and operator that requires the trim to be set to zero after landing, and, 2) that no walkaround was performed (in which case the pilot would surely notice), and 3) that the current trim is outside the allowable takeoff trim range with the to be loaded weight, and 4) that the problem has not been resolved with the before-takeoff checklist completed; I would personally never bother a pilot with this. You're more likely to distract a pilot than fix something covered by checklists and procedures.

@Cloud Would you also inform the pilot that likely, the flaps are not set for takeoff, or the cargo door not secured open, or the engines are shut down, or there are loose objects immediately in front of the engine intake, or or or ...? I've said it before, and I'll say it again: at some point, as passengers, we just have to accept that the professionals paid to do a job know how to do their job. Yes, even professionals sometimes make mistakes, but that doesn't mean that (a) people don't learn from those mistakes, and (b) any given professional will make a mistake the one time you fly.

Looking closely at the photo, that's an Embraer 175.  I Googled other photos of the same model aircraft, and all of them have a similar flat surface.  It appears to be a flat designed into the aft fuselage to accommodate trimming by moving the entire stabilizer ("tail wing") up and down at the leading edge.

Thank you @Zeiss Ikon. I see the Embraer 175 printed now that you mention it.

737's have the same thing.


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Stability and Control of an aircraft
Why is the pitching moment coefficient of an airplane said to be constant about the aerodynamic centre of an aerofoil

You should read how it flies? and search in this website for other questions about pitch stability.

You should edit the title so that it is more specific (less generic) so that it is easier to browse questions about a specific subject (e.g. pitch stability) only reading the titles.

Alright thank you so much

I don't understand what you are after. On one hand you ask about the "moment coefficeint an airplane", but on the other hand you talk about "an aerofoil". An aircraft is a 3D structure which has typically several lifting surfaces, and aerofoil is a 2D cross section of wing. Both have very different aerodynamic behaviour and their aerodynamic centres can't be compared. Please rephase your question to be more clear.

Why is the pitching moment coefficient of an airplane said to be constant about the aerodynamic centre of an aerofoil?

Because the phrase "aerodynamic center" is defined as meaning "the point about which the pitching moment coefficient is constant."
That's it, that's the entire answer. If I've misunderstood the question, please let me know.

Consider this equation:
$$ C_m = C_{m0} + \xi C_L $$
where $C_{m0}$ is $C_m$ at no lift, and $C_m$ is the pitching moment.
At the aerodynamic centre $\xi = 0$ and $C_L = 0$, $$ C_m = C_{m0} $$ which is a constant.

Hi Shishir Maurya. I edited the equation into the answer, as opposed to having it included as an image. Please check to make sure I didn't introduce any errors while doing so. You can find a Mathjax tutorial over here, and this tool is handy for fancy symbols.

Thanks, I am new here. I will surely learn the Mathjax and you edited correctly.


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What is the identity of this abandoned, wrecked Sudan Airways airliner?
The other day, I was browsing Twitter, when I came across this picture of an airliner posted by a Sudanese local (thought it was a Boeing 737, but a friend of mine pointed out that it was an Airbus A300). I tried Google Image Searching it, but couldn't find any more details about how it got to be wrecked in this condition ...
Could anybody shed a light on this?

According to this person the photo is of an Airbus A300B4-620, ST-ASS (Ex F-WZLR, F-ODTK) that is being scrapped at Khartoum Airport. In his photo the cut in the fuselage closest to the cockpit has been made, but the one further aft has not yet. You can see by the shape of the first cut that it is the same aircraft.

It looks to me more like an abandoned plane (that's fallen apart) rather than a crashed/wrecked one. A Sudanese A310 crashed at Khartoum in 2008 - but it caught fire and was destroyed, which clearly hasn't happened here unless someone came and repainted it afterwards.
This list gives ten A300s which have previously been owned by Sudan Airways; TC-OYC is stored at Istanbul (and this doesn't look like Istanbul), ST-ATA is stored location unknown, and ST-ASS was scrapped while last owned by Sudan. Of their A310s, ST-AST, is stored (location unclear); ST-ATN was destroyed by fire after a crash. So that's three A300/310s which might plausibly be lying derelict at a Sudanese airport.
For an alternative count, this thread lists four A300s "stored" at Khartoum, ST-ATA, ST-ATB, ST-ASS & ST-AST; all put up for sale as parts/scrap a few years back but never sold. No pictures but it's clear they're not in a great state.

This is an Airbus A300. Sudan Airways has not had an accident with this type of aircraft and as such it must be a retired plane.

This seems unlikely to be the same aircraft - the Flight 109 investigation report seemed to indicate much more substantial fire damage than is shown in the picture. Is it more likely to be an abandoned/"stored" aircraft than a crashed one?

@Andrew You are right, should have checked other photos before.


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Why are fighter jets equipped with dorsal fin?
I don't really get whats the purpose of dorsal fins and what they do with the flow over the vertical stabilizer. I noticed on most single tail configured aircrafts are equipped with a dorsal fin like the f 16.

The F-16 doesn't have a dorsal fin.. but it does have a largish fairing at the root of the fin. If this is what you're talking about then its function is probably to cover the stab-fuselage joint. It does have ventral fins on the underside which serves to enhance stability.

@Anilv at the front of that fairing joint there is a small fin, which I believe is what the OP is referring to.

Can you add a picture with some markings to help us understand please?

Related: What is the purpose of a dorsal fin such as on the Boeing 737?

There are two features that you may be referring to.
A) Is an antenna that is shaped to provide low drag. Not all F16 have this.
B) The true dorsal "fin":

increases structural strength
decreases drag by providing a fillet for the vertical stabilizer,
improves stability about the vertical axis,
and specifically in the case of the f16 (when installed)
     aerodynamically blends the drag chute pack (box at base of Rudder)
many aircraft use this as a convenient air intake for
     equipment or auxiliary cooling.

source

Adding to point 5, as an F-16 lover, on the A and B models the fin is shorter than in C and D models.
Also, in D and F models there is the option to have a complete dorsal spine, going from the end of the cockpit to the vertical stabilizer, longer or shorter depending on the presence of the drag chute.

If you are asking about the aerodynamic purpose of a dorsal fin, then dorsal fin actually increases the angle of side slip to get maximum side force. At high side slip angles, it produces stable vortices along the surface of the vertical tail which  produces higher suction force over vertical tail, thus increasing the side force.
So in absence of the dorsal fin at high side slip angle, aircraft would have stalled i.e. just as jwzumwalt said, it provides stability along the vertical axis.  According to me as F-16 is a fighter aircraft which requires high maneuverability, it would require a dorsal fin to stabilize at high side slip angles.

The fin acts as a directional stabiliser, but not all aircraft or fighters have them. The Horten IX (Gotha Go 229) was a notable flying wing fighter design that didn't have a fin.


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What effect does downwash have on the horizontal stabilizer
Fighter aircrafts usually have their horizontal stabilizers in lined with the main wing which would subject it to downwash. How does this effect the stability of the aircrafts and the effectiveness of the elevator?

First, the downwash extends over a wide height, so moving the tail up or down will change the local downwash angle only a little (first order approximation: By the cosine of the elevation angle at which the tail is positioned).
Next, downwash at the tail is a good thing: Since the downwash angle increases with angle of attack, the angle of attack range at the tail is smaller by the downwash effect than it is for the wing. This keeps the tail in a narrower band of angles and allows it to still stay firmly in the linear range when the wing is about to leave it (which happens when the wing stalls).
What you don't want to happen is that the wing wake hits the tail. The wake contains the wing's boundary layer and produces an undesirable drop in dynamic pressure at the tail. Initially, tails were moved up to avoid the wing wake impinging on the tail (think MiG-15 or F-101 Voodoo), but that only shifted the eventual point where the wing would blanket the tail to higher angles of attack. With the reduction in aspect ratio and the increase in wing sweep made possible by more powerful engines, the upper angle of attack limit went up and the tail could not be moved up enough. Therefore, all modern designs try to put the tail as low as practical and even add some anhedral. Here the limit is of course a tail strike, so you might get the impression that the tail is in line with the wing. At practical angles of attack it isn't.

Thank you Peter, I understand that blanketing of the vertical tail reduces the effectiveness of the rudder and might produce bad spin resistance. But I am a bit confused to what exactly is blanketing? Is it just turbulent flow over the rudder?

@NareshSabapathy: Yes, and more. Turbulence from the wake of the wing also means lower dynamic pressure, so all forces are much lower. Same goes for control effectiveness: The turbulence in the wake means that all flow is separated and dynamic pressure is much lower than in clean flow.


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Does headwind on takeoff affect g force?
I'm aware that headwind or tailwind does not affect rate of climb or descent but only the angle. I'm not sure if it affects the g force or not. I'd assume it affects g force because using trignometry, using SIN theta the vertical component increases as the angle increases. However a confirmation would be aappreciated.

The question would benefit from just asking about climb angle and leaving takeoff out of it-- as it stands it's not clear whether you are asking about the G-load imposed as we transition from horizontal to climbing, or the G-load in the climb, or both-- though the answer ends up being the same in all cases.

Or for that matter the G-load (horizontal) pushing the pilot back in his seat during the takeoff run-- is this being considered?  See above comment, I think the question can be simplified.

Wait a minute here, please clarify exactly what you mean by "using sine the vertical force component increases as the angle increases."   What is theta?  What vertical component are you saying increases as theta increases?  See diagrams attached to this answer to related question https://aviation.stackexchange.com/questions/40921/does-lift-equal-weight-in-a-climb/56476#56476 .  If by G-load you are just looking at the component of total aerodynamic force that acts in up-and-down direction in a/c reference frame, this will be proportional to L in the diagrams.

(Ctd) It almost sounds like you are suggesting that the G-load is GREATER the steeper the climb angle.  I think this question can use some clarification-- at least explain what you mean by theta and what vertical force component you are saying increases with climb angle.

sorry for the late reply. Yes i had assumed the vertical component is the g force. The theta is using sin theta to calculate the vertical component during takeoff

Well the vert comp of lift + the vert comp of (thrust - drag) just equals weight, if acceleration is zero-- do the diagrams here match your understanding?  -- https://aviation.stackexchange.com/questions/40921/does-lift-equal-weight-in-a-climb/56476#56476 -- I used "c" instead of theta for the climb angle.  It still seems to me that there may be some incorrect ideas built in to the language of the question.  Or are you trying to look at what happens right as we transition from level flight into climb, at which moment there must indeed be a net force acting perpendicular to flight path--

Thanks. The answer from XXavier has cleared it up.

In the takeoff run, the acceleration is less with headwind, but once in flight, the airplane flies within the mass of air. The fact that that mass may move in relation with the ground doesn't affect the magnitude of the forces involved in flight, so accelerations are not affected by the wind.

Tell this to the albatross bird that can gain velocity by flying through a changing wind field. If acceleration wouldn't be affected by the wind speed this would not be possible. (dynamic soaring)

Dynamic soaring is an extremely fascinating phenomenon but has nothing to do with maneuvering within a UNIFORM airmass.  There is a website available that claims otherwise but please let's not contaminate our minds with those thoughts.  Anyone already contaminated by said website may wish to formulate a separate question for Aviation SE on the topic.

No--
For flight at any given angle-of-attack, the aircraft's pitch attitude in space is tied to the climb or descent angle in relation to the airmass, not in relation to the ground.
The climb angle achieved with respect to the airmass does NOT change depending on whether the aircraft is facing upwind or downwind, and therefore the aircraft's pitch attitude in space doesn't depend on whether the aircraft is facing upwind or downwind.
Just as a glider's pitch attitude in space doesn't change as it circles with a given angle-of-attack and airspeed, even in the presence of a very strong tailwind  that drops its groundspeed to zero at times.
So even if we recognize that the component of the G-load that acts in the "up and down" direction in the aircraft's reference frame is reduced when the aircraft is in a nose-high (or nose-low) pitch attitude, we won't see a difference in this value when we climb upwind versus downwind.
Note-- it is a bit ambiguous as to exactly what "G-load" means.  Is it what we read on the G-meter-- i.e. just the component of "felt" acceleration that acts in the up-and-down direction in the aircraft's reference frame?  If so, this is just equal to the component of the net aerodynamic force that acts in the up-and-down direction in the aircraft's reference frame.  Essentially, the magnitude of the lift vector, divided by the aircraft weight.  The steeper the climb angle, the smaller the lift vector -- see related answer Does lift equal weight in a climb? .
Or by G-load, do we mean the total "felt" acceleration, including the component that acts in the fore-and-aft direction in the aircraft's reference frame?  If so, this is just equal to the net aerodynamic force the aircraft is generating, divided by the aircraft weight. Since in a stabilized climb with constant airspeed and constant direction of flight path, the net aerodynamic force generated by the aircraft IS exactly equal to weight, the G-loading by this definition would always be "1" in a stabilized climb, regardless of climb angle.
At any rate, by neither definition of "G-loading" do we see a difference when climbing upwind versus when climbing downwind.  Nor do we see a difference in the aircraft's pitch attitude.
(Nuances-- this answer assumes that EITHER the pilot and G-meter are located at the aircraft CG in the fore-and-aft sense, or the pitch rotation rate is zero.  Otherwise the relationship between G-meter reading (and "felt" acceleration) and aerodynamic force is influenced by the pitch rotation rate, as has been recently been pointed out in comments to other related answers.  But even considering such added complications, no difference is caused by climbing upwind versus downwind.)

Note that as we initially pitch up into the climb, there's an INCREASE in G-load-- we must exert a "centripetal" force to make the flight path curve skywards.  For typical light planes this extra force is small, but it cannot be zero.  That didn't seem to be what you were asking about, and it would not be changed depending on whether we were taking off w/ headwind versus tailwind, for a given airspeed and a given rate of pitching up into the climb.

An airplane flying into a headwind and pitching up will feel the same G As flying in stationary air, but the angle of climb will increase.
It will increase with the ratio of arccos (V_plane - Vwind)/V_p.

I assume you are not trying to suggest that there will be any change in the aircraft's pitch attitude?

@quietflyer. No change in pitch flying into a headwind even if the plane is accelerating. Same pitch as normal take off pitch. That's why we land or take off into the wind keeping the same pitch, unless we have cross wind.

Key here is to realize that head wind, tail wind, climb, descent, Wright Flyer, sailplane, jet, balloon, the physics is the same regarding acceleration and non-accelerated flight.  The four forces, simple as they seem, can be combined in an infinite number of orientations to achieve zero acceleration.
This is NOT necessarily just standing still, it can be an equilibrium of forces at a given velocity and direction.  This, obviously but critically, applies to heavier than air flight. "If you ain't moving, you ain't flying". There for, in steady state flight, regardless of orientation or velocity, G force is only from gravity, and it is 1.

When you are only looking for the static acceleration that the pilot feels in his/her seat then the g-force is only a factor of the pitch attitude (theta). That is if we assume a stationary climb or descent, which by definition has a netto acceleration of zero. So 1g pulling the aircraft towards the ground, the more pitch attitude we have the more of the gravity has to be compensated by engine thrust to maintain an unaccelerated flight. This leaves less a smaller vertical g component. At 90deg pitch up the vertical component (from your pilot view) is going to be zero but the acceleration pushing you into the seat is now 1g (and engine thrust must equal aircraft weight). So in your fighter jet at 90 degrees nose up you will just feel as if you are lying on your back and you don't feel any vertical acceleration anymore, regardless of the wind speed.

Ergo the vertical acceleration felt by the pilot in stationary flight is
cos( theta ) * g, longitudinal acceleration on the pilot is sin(
theta ) * g.

Comments are not for extended discussion; this conversation has been moved to chat.


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What is being depicted w.r.t. AoA in this video?

I was looking at this video of the AF447 crash simulation and at about 4:45 the angle of attack diagram shows a few things I don't understand (frame above).
I get the orientation of the airplane, pointing a few degrees above the horizon, but what does the green line represent in layman's terms? Is that the direction the plane needs to go for the wings to generate lift?
Also, I see the red line has to do with the warning, but how does that play into it as well?

Can you clarify if you're asking about the line in the upper-right inset or something else?

There are several colored lines.  Which one are you asking about?  I'm assuming you are referring to the (cyan?) and red lines in the upper right inset.

The cyan line shows the current direction of the airflow.  When compared with the orientation of the airplane can show the angle of attack.  At this time, the relative wind is coming at a very steep angle to the airplane and the wing is stalled.
Another way to see it is that reversing this arrow shows the direction the airplane is moving through the air.  In a no-wind situation, this would also be the direction the plane is moving with respect to the ground.
The red line is at a constant angle with respect to the airplane and shows how low the airflow can go without causing a warning.  As it is below this line, the stall warning has sounded.

Maybe it would be good to add a sidenote that the cyan line shows the opposite of „where the plane goes“ (I assume that for non-aviation folks earth fixed is the more natural frame of reference rather than airplane fixed).


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Detecting faults in CVR mic inputs
Years ago my father worked for Eastern Airlines out of Miami (that should give you a timeframe) and he was an electrical engineer.  I recall once he told me that the pilots (one or more) used to intentionally damage the CVR microphone in the cockpit by twisting it until the contacts broke.  Their reasons are not relevant for this question.
My question is, knowing what model planes they flew back then, (727, Tristars, A300s, 757, etc) how did the mechanics know that the mic was not working?  Would the system show it as a fault or did the CVR tape review show that there was no data from that mic?  The former I'd think would allow them to track down the pilot(s) that were responsible for this act.
Also, on planes with more modern electronics, would this fault be detectable immediately or at least before the next flight?

Interesting question. On the Boeing 727 there is a panel named "Cockpit Voice Recorder Microphone Monitor". A headphone can be plugged into the labelled jack to test if the cockpit voice recorder microphone is working properly. I have not checked the other model aircraft you mentioned but I assume they have the same test capability.

These microphones are usually passive devices. I doubt there is a warning system even on modern airliners to tell you the CVR mike is not working. But I would also be willing to bet good money they are no longer so easy to disable.

great info, thanks.  Seems like a lot of surface area to test such a minute (yet important) function.  Do you know if its a scheduled maintenance procedure to test this, part of an A check perhaps?

I know that the checks are recommended to be performed regularly, but my guess is that each airline has a different schedule to do so.

@JuanJimenez: Just recommended?  Not required?

The FARs do not require the microphone itself to be tested on a regular basis, not that i can see.


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Is it the tower's purview to recall an aircraft from the runway for loading the wrong cargo?
On August 17, Ryanair flight FR 11 entered the runway after already being delayed for 1.5 hours, only to exit at the first available opportunity. After being jokingly welcomed to Stansted from where we had very nearly departed, we proceeded to wait for at least 1.5 hours more while the wrong cargo was being unloaded to make way for the correct one.
The crew told us the delay was due to having loaded the wrong cargo, and made it very clear to us that this wasn't their fault, which I have no reason to doubt. The pilot made it sound like the tower told them to abort. We were already on the runway and turned back at the very last minute.
My question is this: Does the tower have the authority to tell an aircraft to abort take-off for having loaded the wrong cargo?
My impression is that the tower is a traffic cop and safety officer rolled into one, and operational questions such as which plane does this cargo belong are outside their normal range of concerns.

Misloaded cargo is a safety issue, because if the weight differs from what was calculated on the weight and balance paperwork, the plane could be overloaded or out of trim (balance). Results could range from nothing to a fatal accident. Other concerns are excess quantities or prohibited hazardous materials being carried on a passenger flight.

For what it's worth I did submit a case with Ryanair who have compensated me to the statutory amount without challenging my claim.

Yes, the tower can tell an aircraft to abort take off and instruct it to return to the gate, although it probably never got that far in this case, I'm guessing that they got to the runway or were asked to line up when the request came in. This was probably at the direction of Ryanair, since the tower would otherwise not have any information on the wrong cargo being loaded, nor would it really care.
I'm guessing at this part....
The back-and-forth probably went like this:

Tower: Ryanair 1234 line up and wait
  Ryanair: Line up and wait, Ryanair 1234
  Tower: Ryanair 1234 taxi on runway to Delta, call company
  Ryanair: Taxi delta, call company

The tower probably didn't relate too much information about what is wrong, the pilots have the ability to talk to the dispatcher for Ryanair to get more information.
Regardless, a tower controller can tell an aircraft that a take-off clearance is rejected, even after the take-off run begins. It is up to the pilot to determine if they can safely reject the take-off and to follow those instructions.
The probability though that the tower knew that the incorrect cargo was loaded is probably zero, at least until somebody from Ryanair called them and had them stop the take off. I'm guessing that they didn't actually get take-off clearance (line up and wait is not take-off clearance) and either they were asked to line up, or they were waiting to enter the runway before they were told to go back (and taxiing on the runway is the easiest way to turn around).
Typically Tower will not call a RTO unless there is a safety concern like an incursion or they notice a wing fell off. Once clearance is given it is rarely rejected. If they had gotten clearance Tower would probably have advised them to call company "once in the air" rather than reject a take off.
If you want to know what it sounds like for Tower to reject a take-off, listen to this YouTube about a near miss regarding similar call signs and two aircraft taking off at the same time.

And as a comment, it looks like that particular flight is very consistently delayed up to 2 hours or more.

Your example comms is very plausible, but I think it's important to emphasize that the answer to the question as literally asked is "no": although ATC can instruct an aircraft to abort take-off, they wouldn't do so for a non-safety reason. What you've described, where they interrupt departure checks to request they call company, is far more likely. It's also possible tower wasn't involved and the aircraft heard from Ryanair directly via some other communication method.

@DanHulme You are right in that it can't (probably) tell an aircraft to go back because of incorrect cargo, but as you said the tower probably wouldn't care or have that information anyway, so I'm sure some panicked loadmaster after noticing the wrong container gave a call up to tower and told them not to let them take off. It is surprising that the aircraft was able to be loaded, go through clearance delivery, push back, ground control and finally take the runway before they noticed the error.

@DanHulme That's important I think. I intend to claim for the delay on that flight and their little excuse of wasn't our fault just doesn't do it for me. They chose to abort, for one reason or another, they weren't forced to.

@rath I don't think it's at all relevant in terms of claiming for the delay, but that aspect is outside our topic area. You can ask on [travel.se] if you want more details about the passenger side of air travel.

@rath Depending on what was in the cargo, it may have been illegal to transport it to the country you were flying to, so it is plausible that it wasn't their fault (at least not the pilots), but somebody at Ryanair loaded the wrong cargo.

@DanHulme I fully intend to, which is why I limited this question to the "technical" aspect :)

It could also be possible for the cargo to have a safety effect: e.g. if the weight of the loaded cargo was significantly more than the planned weight, the aircraft could be over MTOW or out of CG limits.

Nevermind the CG and MTOW, all performance calculation are out of whack... rotation speeds, stop distance... a safe takeoff is not guaranteed if you put the wrong cargo in the hold

"This was probably at the direction of Ryanair, since the tower would otherwise not have any information on the wrong cargo being loaded, nor would it really care." Exactly. The tower doesn't force the airplane to stand down, there's no safety risk after all. They're merely the messenger.

There is also a slight formal difference between a "cancelled" and a "rejected" clearance.

@Mast There are plenty of comments above yours explaining that loading the wrong cargo is most definitely a safety issue.

@Radu094: And that's even assuming that the cargo that shouldn't have been loaded doesn't contain anything inherently dangerous.

There's no way to know for sure, since UK does not allow us to listen to ATC transmissions, BUT...
I seriously doubt the tower told them to abort the takeoff. I just don't see the company calling the tower unless it's a safety related problem. The company almost certainly called the aircraft who then had to inform the tower they had to return to the gate and request taxi instructions.
For the company to call the tower for an operational situation would most likely not be acceptable and it would involve an extra step anyway. They would call the aircraft directly, either over the radio or through an ACARS message.
Not the pilots' fault, nor ATC's fault, but the airline's fault, for sure.


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Why are helicopters less efficient than planes?
At higher speeds, advancing helicopter blades will operate at velocities approaching mach 1, reducing efficiency. But what about lower speeds?
A helicopter is basically a wingless plane with a giant propeller. Why is it more efficient to move around using propellers to provide forward trust while using separate fixed wings to provide lift?

They don't fly so much as beat the air into submission...

...by using money ;-).

The answer lies in the difference between energy and momentum. Both have to be maintained, but energy goes up as the square of velocity, while momentum goes up linearly. To impart a given momentum, you can either move a lot of air slowly, or a little air quickly -- but as you move the air more quickly, it requires disproportionately more energy.
In other words, the bigger the wing, the more efficient (from an energy/momentum perspective, anyway).
A rotor's size is limited, both because its fast rotation means it needs to be light (and thus can't be very reinforced, etc) and because the larger its radius, the faster the end spins. That means it has to touch less air, but move it more -- which requires more energy.
There's more detail at this question on Physics StackExchange (disclaimer: I wrote that question).

Thanks for the link to your question! It's basically what I was asking, but better elaborated. I like the third answer to your question.

I don't see why the end has to spin faster as the radius is increased. In fact, it seems to me like the opposite should be true. As the radius is increased, the wings of the rotor are longer, so they don't need to move through the air as fast to provide the same lift.

@StephaneBersier That's getting a bit too far from my expertise. :-) But I wonder if the problem is that if you did that, then the inner parts of the rotor would be moving too slowly. There are basically a lot of "radius-squared" tradeoffs to be made. Beyond that, just the physical aspect of making such a wing that's both very long and very lightweight (it does have to move in a lot of different ways, so you want to keep its inertia down) is hard.

A helicopter is not exactly wingless - the rotor is the helicopter's wings.
A wing's drag is proportional to $W\cdot(L/D)$, where $W$ is weight and $L/D$ is lift/drag. The energy spent overcoming this drag each second is $V_{wing}\cdot W\cdot(L/D)$, where $V_{wing}$ is the wing's airspeed.
The amount of fuel required to travel distance $S$ is then $SFOC\cdot S/V_{aircraft}\cdot V_{wing}\cdot W\cdot(L/D)$, where $V_{aircraft}$ is the aircraft's ground speed, where $SFOC$ is specific fuel consumption.
For a fixed wing, $V_{wing}$ and $V_{aircraft}$, ignoring wind, are the same. In a rotary wing (helicopter), $V_{wing}$ is higher than $V_{aircraft}$ - the wing travels a longer path through the air, with rotation.
For this reason, a helicopter will always be less efficient than a fixed wing aircraft with the same lift/drag ratio. If the ratio is 3x, that's 3 times more fuel for the same distance. This is of course a vast simplification, ignoring all the details and just describing the principle.
To put it in even simpler terms, a helicopter is like a plane with its wings flying a path full of loops.

A nice and well-founded answer...

I feel like your answer is missing something, as it doesn't apply the same math to the propeller of the plane, which is moving even faster than the rotor of a helicopter... sadly I don't have more time to think about it for now.

@StephaneBersier A propeller's thrust is much smaller than that of a rotor, just ~1/10 of the aircraft's weight. The wings then turn that thrust into 10 times more lift, to keep the plane aloft. Since propellers are fairly small, their drag is also relatively small. Notable fact: helicopters use the most fuel per minute in hover; going faster takes less fuel, because the rotor no longer needs to provide entirely static thrust.

the wing on a small plane moves through the air at around 100MPH, whereas the main rotor blade on a small helo moves through the air at around 400MPH. the blade is smaller than the wing to be sure, but the drag forces acting at 400MPH are still big, and soak up a lot of horsepower.
Try this comparison: imagine a Cessna 150 with two people in it at cruise conditions and compare it with a Robison R22 with two people in it, same altitude, etc. Look at their respective fuel burn rates.

What about the speed of the propeller blades on the Cessna 150? I'd imagine it's above 180 m/s.

yes, but the prop diameter on a C150 is quite small compared to the main rotor of an R22. The Cessna cruises at about 110mph on 100HP and burns between 5 and 5.6 GPH. The R22 requires 150HP and burns between 7.5 and 10 GPH.

As pointed out in this answer:

More interference drag
A circular shape of the rotor disk - it can be considered to be a wing in forward flight, but a circular wing is much less efficient than a long slender wing.

The picture from J. Gordon Leishman, Principles of Helicopter Aerodynamics, illustrates the complexity of the multitude of aerodynamic drags that can be found in a helicopter, and that decrease efficiency.